Chemistry Q71 – JEE-MAIN 2026

Home Chemistry Chemical BondingQuestion #223

Consider the following species:

B r F_{5}

X e F_{5}^{-}

B F_{4}^{-}

I C l_{4}^{-}

X e F_{4}

S F_{4}

N H_{4}^{+}

C l F_{3}

X e F_{2}

I C l_{2}^{-}

. Number of species having

s p^{3} d

hybridized central atom is _______.

Enter Calculated Numerical Value:

Clue 1 of 3: Hybridization Concept

Hybridization of the central atom is determined by its steric number, which is the sum of sigma bonds and lone pairs.

🧠 Full Solution Path

Step 1: Calculate Steric Number for Each Species✦ Active

The steric number (SN) for the central atom in each species is calculated as the sum of the number of sigma bonds (bond pairs, BP) and the number of lone pairs (LP) around it. The number of valence electrons on the central atom, adjusted for charge, is used to determine lone pairs.

\begin{array}{lclcl} Species & {Valence e}^{-} & BP & LP & SN \\ B r F_{5} & 7 & 5 & (7 - 5) / 2 = 1 & 6 \\ X e F_{5}^{-} & 8 + 1 = 9 & 5 & (9 - 5) / 2 = 2 & 7 \\ B F_{4}^{-} & 3 + 1 = 4 & 4 & (4 - 4) / 2 = 0 & 4 \\ I C l_{4}^{-} & 7 + 1 = 8 & 4 & (8 - 4) / 2 = 2 & 6 \\ X e F_{4} & 8 & 4 & (8 - 4) / 2 = 2 & 6 \\ S F_{4} & 6 & 4 & (6 - 4) / 2 = 1 & 5 \\ N H_{4}^{+} & 5 - 1 = 4 & 4 & (4 - 4) / 2 = 0 & 4 \\ C l F_{3} & 7 & 3 & (7 - 3) / 2 = 2 & 5 \\ X e F_{2} & 8 & 2 & (8 - 2) / 2 = 3 & 5 \\ I C l_{2}^{-} & 7 + 1 = 8 & 2 & (8 - 2) / 2 = 3 & 5 \end{array}

Step 2: Identify Hybridization from Steric Number○ Expand

The hybridization of the central atom is determined by its steric number (SN):

\begin{array}{lcl} SN & Hybridization \\ 4 & s p^{3} \\ 5 & s p^{3} d \\ 6 & s p^{3} d^{2} \\ 7 & s p^{3} d^{3} \end{array}

We are looking for species with $s p^{3} d$ hybridization, which corresponds to a steric number of 5.

💡 Teacher's Secret Hint

Remember that the number of sigma bonds is equal to the number of atoms directly bonded to the central atom.

Step 3: Count Species with

s p^{3} d

Hybridization○ Expand

From the calculations in Step 1, the species with a steric number of 5 are:

1. $S F_{4}$ (SN = 5, $s p^{3} d$ )

2. $C l F_{3}$ (SN = 5, $s p^{3} d$ )

3. $X e F_{2}$ (SN = 5, $s p^{3} d$ )

4. $I C l_{2}^{-}$ (SN = 5, $s p^{3} d$ )

Therefore, there are 4 species with $s p^{3} d$ hybridized central atoms.

💡 Teacher's Secret Hint

Ensure you correctly count the lone pairs, especially for polyatomic ions where the charge affects the total valence electrons.

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Chemistry Question 71 – JEE-MAIN 2026

Clue 1 of 3: Hybridization Concept

🧠 Full Solution Path